Calculation verification please

Not possible. With a 1:40 SUT into 47000 ohms, the cartridge will see a -30 ohm load. The sum resistance of two parallel resistors cannot exceed the value of the lowest value resistor. So any resistor you place in parallel will reduce the impedance, not augment it. But if your cartridge has internal resistance of 1.4 ohms, it will drive 30 ohms. So why do you want 375 ohms?

I’ve been experimenting into a MC section and the manufacturer says to load between 100 - 800 ohms. I’ve tried 250, 324, 368, 404, 500 & 600 ohms and found I like best the 368 ohm loading. So now want to try using my SUT. Yes it will be fine going directly into the 47K MM section but thought I’d like to try loading it to see around the 375 ohm load.

ChatGPT said:

To determine the load the cartridge sees through the Step-Up Transformer (SUT), we use the following formula:

Rcartridge=Rphonoturns ratio2R_{ ext{cartridge}} = rac{R_{ ext{phono}}}{ ext{turns ratio}^2}Rcartridge​=turns ratio2Rphono​​

Given:

• Turns Ratio = 1:40
• Phono Input = 47,000Ω (47kΩ)

Rcartridge=47000402=470001600=29.375ΩR_{ ext{cartridge}} = rac{47000}{40^2} = rac{47000}{1600} = 29.375ΩRcartridge​=40247000​=160047000​=29.375Ω

So, the cartridge sees a 29.4Ω load.

Adjusting the Load to 375Ω

To achieve a 375Ω load at the cartridge, we need to calculate the required parallel loading resistor at the phono stage.

The formula for the effective load seen by the cartridge when a parallel resistor (RLR_LRL​) is added to the phono input is:

Rcartridge=RL∥Rphonoturns ratio2R_{ ext{cartridge}} = rac{R_L parallel R_{ ext{phono}}}{ ext{turns ratio}^2}Rcartridge​=turns ratio2RL​∥Rphono​​

First, solve for the necessary parallel resistance:

RL∥47000=375×1600R_L parallel 47000 = 375 imes 1600RL​∥47000=375×1600 RL∥47000=600,000ΩR_L parallel 47000 = 600,000ΩRL​∥47000=600,000Ω

Using the parallel resistor formula:

Rparallel=RL×47000RL+47000R_{ ext{parallel}} = rac{R_L imes 47000}{R_L + 47000}Rparallel​=RL​+47000RL​×47000​

Setting this equal to 600,000Ω:

RL×47000RL+47000=600000rac{R_L imes 47000}{R_L + 47000} = 600000RL​+47000RL​×47000​=600000

Solving for RLR_LRL​:

RL×47000=600000(RL+47000)R_L imes 47000 = 600000 (R_L + 47000)RL​×47000=600000(RL​+47000) RL×47000=600000RL+28200000000R_L imes 47000 = 600000 R_L + 28200000000RL​×47000=600000RL​+28200000000 RL(47000−600000)=28200000000R_L (47000 - 600000) = 28200000000RL​(47000−600000)=28200000000 RL(−553000)=28200000000R_L (-553000) = 28200000000RL​(−553000)=28200000000 RL=28200000000553000≈51.0kΩR_L = rac{28200000000}{553000} approx 51.0kΩRL​=55300028200000000​≈51.0kΩ

So, to achieve a 375Ω load at the cartridge, you need to add a 51kΩ resistor in parallel with the 47kΩ phono input.

My phono unit has the ability, via RCA plugs, to load either the MC or MM section. Very easy and simple. 

I refuse to read the gibberish, but if you place a 51K resistor in parallel with a 47K resistor, you will get around 24K resistance. If you use your 1:40 SUT into 24K resistance, you will end up with a phono load of about 15 ohms. In other words, you’re going backwards. I don’t care what the book said. But your cartridge can drive even 15 ohms. So if you do the 51K mod, it will work ok, but the load is not 375 ohms. The only way to increase the resistance of a resistor is to place another resistor in series with that resistor, and obviously you don’t want to do that in this case.

just for the fun of it, if you have an ohm meter, and if you have two 50 K resistors, place them in parallel and measure the resistance across the parallel pair. It will be 25K ohms.

what you could do is replace the 47K load resistance with a 600K load resistor. Then using your 1:40 SUT, the cartridge will see 375 ohms load.